<h3>
Answer: 3^15</h3>
Work Shown:
3^12 * 27
3^12 * 3^3
3^(12+3)
3^15
I used the rule a^b*a^c = a^(b+c) on the third step.
(m+2)(m+3)= (m+2)(m-2)
⇒ m^2+ 3m+ 2m+ 6= m^2 -4
⇒ 5m+ 6= -4 (m^2 on both sides cancels out)
⇒ 5m= -4-6
⇒ 5m= -10
⇒ m= -10/5
⇒ m= -2
The final answer is m=-2~
Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be
.
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.


Amount of interest earned from lower-yielding account:
.
Amount of interest earned from higher-yielding account:
.

Let us solve for x.



Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be
.
Therefore, the man invested $60,000 at 6%.